We are going to use conservation of angular momentum for this problem. We are going to need the moment of inertia for a rod rotating about its end. I will derive it here using the formula for the inertia about an objects center of mass.

`I_(cm)=int r^2 dm`

Where `r` is...

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We are going to use conservation of angular momentum for this problem. We are going to need the moment of inertia for a rod rotating about its end. I will derive it here using the formula for the inertia about an objects center of mass.

`I_(cm)=int r^2 dm`

Where `r` is the distance from the axis of rotation.

The moment of inertia through the center of mass for a rod on the x-axis with density `lambda=M/L` is:

`I_(cm)=int x^2 dM=int_(-L/2)^(L/2) x^2 (lambda*dx)=lambda/3 x^3|_(-L/2)^(L/2)=1/12 lambda*L^3=12 ML^2`

The parallel axis theorem states the the moment of inertia about the end of the rod will then be:

`I_(rod)=I_(cm)+Mr^2`

Where `r` is the distance from the center of mass.

`I_(rod)=1/12 ML^2+M(L/2)^2=1/3ML^2`

Now for conservation of momentum. Let the quantities with the subscript c be for the clay.

`L_i=L_f`

`r xx p_c=I_(t o t) omega`

`omega=(r xx p_(c))/I_(t o t)`

`omega=(L*mv*sin(90^@ -alpha))/(I_(rod)+I_(c))`

`omega=(L*mv*cos(alpha))/(1/3ML^2+mL^2)`

`omega=(3mv*cos(alpha))/((M+3m)L)`

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